Answer: [tex]0.604 m/s^{2}[/tex]
Explanation:
The complete question is:
Mary and Sally are in a foot race (see the figure). When Mary is 22 m from the finish line, she has a speed of4.0 m/s and is 5.0 m behind Sally, who has a speed of5.0 m/s. Sally thinks she has an easy win and so, during the remaining portion of the race, decelerates at a constant rate of 0.44m/s2 to the finish line.
What constant acceleration does Mary now need during the remaining portion of the race, if she wishes to cross the finish line side-by-side with Sally?
We have the following data:
Mary's speed: [tex]V_{M}=4m/s[/tex]
Mary's distance to the finish line: [tex]d_{M}=22 m[/tex]
Sally's speed: [tex]V_{S}=5m/s[/tex]
Sally's distance to the finish line: [tex]d_{S}=22 m- 5 m=17 m[/tex]
Sally's acceleration: [tex]a_{S}=-0.44 m/s^{2}[/tex] because we are told she decelerates
We need to find Mary's acceleration: [tex]a_{M}[/tex]
But firstly, we have to find the time [tex]t[/tex] it takes Sally to reach the finish line, which is the same time Mary will have to reach the finish line. This can be done by the following equation:
[tex]d_{S}=V_{S}t+\frac{1}{2}a_{S}t^{2}[/tex] (1)
Rewritting the equation:
[tex]\frac{1}{2}a_{S}t^{2}+V_{S}t-d_{S}=0[/tex] (2)
[tex]\frac{1}{2}(-0.44 m/s^{2})t^{2}+5m/s t-17m=0[/tex] (3)
Multiplying by 1 each side of the equation:
[tex](0.22 m/s^{2})t^{2}-5m/s t+17m=0[/tex] (4)
This is a quadratic equation (also called equation of the second degree) of the form [tex]at^{2}+bt+c=0[/tex], which can be solved with the following formula:
[tex]t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}[/tex] (5)
Where:
[tex]a=0.22[/tex]
[tex]b=-5[/tex]
[tex]c=17[/tex]
Substituting the known values:
[tex]t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(0.22)(17)}}{2(0.22)}[/tex] (6)
Solving (6) we take the smaller result (because the longer result is equivalent to the time it takes to Sally to reach the finish line, change her direction and cross the finish line again):
[tex]t=4.18 s[/tex] (7) Â Now that we have the time, we can calulate Mary's acceleration
[tex]d_{M}=V_{M}t+\frac{1}{2}a_{M}t^{2}[/tex] (8)
Isolating [tex]a_{M}[/tex]:
[tex]a_{M}=\frac{2(d_{M}-V_{M}t)}{t^{2}}[/tex] (9)
[tex]a_{M}=\frac{2(22m-(4 m/s)(4.18 s))}{(4.18 s)^{2}}[/tex] (10)
Finally:
[tex]a_{M}=0.604 m/s^{2}[/tex] (11) This is the acceleration Mary needs if she wishes to cross the finish line side-by-side with Sally.
